# Homework 2: Higher Order Functions

*Due by 11:59pm on Thursday, February 6*

## Instructions

Download hw02.zip. Inside the archive, you will find a file called
hw02.py, along with a copy of the `ok`

autograder.

**Submission:** When you are done, submit with ```
python3 ok
--submit
```

. You may submit more than once before the deadline; only the
final submission will be scored. Check that you have successfully submitted
your code on okpy.org. See Lab 0 for more instructions on
submitting assignments.

**Using Ok:** If you have any questions about using Ok, please
refer to this guide.

**Readings:** You might find the following references
useful:

**Grading:** Homework is graded based on
correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus.
**This homework is out of 2 points.**

The `construct_check`

module is used in this assignment, which defines the
function `check`

. For example, a call such as

`check("foo.py", "func1", ["While", "For", "Recursion"])`

checks that the function `func1`

in file `foo.py`

does *not* contain
any `while`

or `for`

constructs, and is not an overtly recursive function (i.e.,
one in which a function contains a call to itself by name.) Note that this restriction does not apply to all problems in this assignment. If this restriction applies, you will see a call to `check`

somewhere in the problem's doctests.

# Required questions

Several doctests refer to these functions:

```
from operator import add, mul, sub
square = lambda x: x * x
identity = lambda x: x
triple = lambda x: 3 * x
increment = lambda x: x + 1
```

### Q1: Product

The `summation(n, f)`

function from the higher-order functions lecture adds
up `f(1) + ... + f(n)`

. Write a similar function called `product`

that
returns `f(1) * ... * f(n)`

.

```
def product(n, f):
"""Return the product of the first n terms in a sequence.
n -- a positive integer
f -- a function that takes one argument to produce the term
>>> product(3, identity) # 1 * 2 * 3
6
>>> product(5, identity) # 1 * 2 * 3 * 4 * 5
120
>>> product(3, square) # 1^2 * 2^2 * 3^2
36
>>> product(5, square) # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
14400
>>> product(3, increment) # (1+1) * (2+1) * (3+1)
24
>>> product(3, triple) # 1*3 * 2*3 * 3*3
162
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q product`

### Q2: Accumulate

Let's take a look at how `summation`

and `product`

are instances of a more
general function called `accumulate`

:

```
def accumulate(combiner, base, n, f):
"""Return the result of combining the first n terms in a sequence and base.
The terms to be combined are f(1), f(2), ..., f(n). combiner is a
two-argument commutative, associative function.
>>> accumulate(add, 0, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19
>>> accumulate(lambda x, y: 2 * (x + y), 2, 3, square)
58
>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
16
"""
"*** YOUR CODE HERE ***"
```

`accumulate`

has the following parameters:

`f`

and`n`

: the same parameters as in`summation`

and`product`

`combiner`

: a two-argument function that specifies how the current term is combined with the previously accumulated terms.`base`

: value at which to start the accumulation.

For example, the result of `accumulate(add, 11, 3, square)`

is

`11 + square(1) + square(2) + square(3) = 25`

Note: You may assume that

`combiner`

is associative and commutative. That is,`combiner(a, combiner(b, c)) == combiner(combiner(a, b), c)`

and`combiner(a, b) == combiner(b, a)`

for all`a`

,`b`

, and`c`

. However, you may not assume`combiner`

is chosen from a fixed function set and hard-code the solution.

After implementing `accumulate`

, show how `summation`

and `product`

can both be
defined as simple calls to `accumulate`

:

```
def summation_using_accumulate(n, f):
"""Returns the sum of f(1) + ... + f(n). The implementation
uses accumulate.
>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> from construct_check import check
>>> # ban iteration and recursion
>>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
... ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
def product_using_accumulate(n, f):
"""An implementation of product using accumulate.
>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> from construct_check import check
>>> # ban iteration and recursion
>>> check(HW_SOURCE_FILE, 'product_using_accumulate',
... ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

```
python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate
```

### Q3: Make Repeater

Implement the function `make_repeater`

so that `make_repeater(h, n)(x)`

returns
`h(h(...h(x)...))`

, where `h`

is applied `n`

times. That is,
`make_repeater(h, n)`

returns another function that can then be applied to
another argument. For example, `make_repeater(square, 3)(42)`

evaluates to
`square(square(square(42)))`

.

```
def make_repeater(h, n):
"""Return the function that computes the nth application of h.
>>> add_three = make_repeater(increment, 3)
>>> add_three(5)
8
>>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> make_repeater(square, 2)(5) # square(square(5))
625
>>> make_repeater(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> make_repeater(square, 0)(5) # Yes, it makes sense to apply the function zero times!
5
"""
"*** YOUR CODE HERE ***"
```

For an extra challenge, try defining

`make_repeater`

using`compose1`

and your`accumulate`

function in a single one-line return statement.

```
def compose1(h, g):
"""Return a function f, such that f(x) = h(g(x))."""
def f(x):
return h(g(x))
return f
```

Use Ok to test your code:

`python3 ok -q make_repeater`

## Submit

Make sure to submit this assignment by running:

`python3 ok --submit`

# Just for fun Question

This question is out of scope for 61a. Do it if you want an extra challenge or some practice with HOF and abstraction!

### Q4: Church numerals

The logician Alonzo Church invented a system of representing non-negative integers entirely using functions. The purpose was to show that functions are sufficient to describe all of number theory: if we have functions, we do not need to assume that numbers exist, but instead we can invent them.

Your goal in this problem is to rediscover this representation known as *Church
numerals*. Here are the definitions of `zero`

, as well as a function that
returns one more than its argument:

```
def zero(f):
return lambda x: x
def successor(n):
return lambda f: lambda x: f(n(f)(x))
```

First, define functions `one`

and `two`

such that they have the same behavior
as `successor(zero)`

and `successsor(successor(zero))`

respectively, but *do
not call successor in your implementation*.

Next, implement a function `church_to_int`

that converts a church numeral
argument to a regular Python integer.

Finally, implement functions `add_church`

, `mul_church`

, and `pow_church`

that
perform addition, multiplication, and exponentiation on church numerals.

```
def one(f):
"""Church numeral 1: same as successor(zero)"""
"*** YOUR CODE HERE ***"
def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
"*** YOUR CODE HERE ***"
three = successor(two)
def church_to_int(n):
"""Convert the Church numeral n to a Python integer.
>>> church_to_int(zero)
0
>>> church_to_int(one)
1
>>> church_to_int(two)
2
>>> church_to_int(three)
3
"""
"*** YOUR CODE HERE ***"
def add_church(m, n):
"""Return the Church numeral for m + n, for Church numerals m and n.
>>> church_to_int(add_church(two, three))
5
"""
"*** YOUR CODE HERE ***"
def mul_church(m, n):
"""Return the Church numeral for m * n, for Church numerals m and n.
>>> four = successor(three)
>>> church_to_int(mul_church(two, three))
6
>>> church_to_int(mul_church(three, four))
12
"""
"*** YOUR CODE HERE ***"
def pow_church(m, n):
"""Return the Church numeral m ** n, for Church numerals m and n.
>>> church_to_int(pow_church(two, three))
8
>>> church_to_int(pow_church(three, two))
9
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

```
python3 ok -q church_to_int
python3 ok -q add_church
python3 ok -q mul_church
python3 ok -q pow_church
```